51nod1394 Differences and problems

I will only use line tree to write ... Do not like tree arrays: Actually, it's not too slow? Then all kinds of * when forget Longlong has been wa ... Pills!

And I don't know how to use map discretization ... So, Sort+unique.

#include <cstdio> #include <cstring> #include <cctype> #include <algorithm>using namespace std; #define REP (I,s,t) for (int. i=s;i<=t;i++) #define DWN (i,s,t) for (int i=s;i>=t;i--) #define CLR (x,c) memset (X,c, sizeof (x)) #define LL long long#define Lson l,mid,x<<1#define Rson mid+1,r,x<<1|1int Read () {int X=0;char c= GetChar (); while (!isdigit (c)) C=getchar () and while (IsDigit (c)) x=x*10+c-' 0 ', C=getchar (); return x;} const int Nmax=1e5+5;const int INF=0X7F7F7F7F;LL sm[nmax<<3],col[nmax<<3];int A[nmax<<1],b[nmax <<1],ct[nmax<<1];struct node{int u,v;}; Node ns[nmax];void update (int p,int ad,int bd,int l,int r,int x) {if (l==r) {Col[x]+=ad,sm[x]+=bd;return;} int mid= (L+R) >>1;p<=mid?update (P,ad,bd,lson): Update (P,ad,bd,rson); sm[x]=sm[x<<1]+sm[x<<1|1 ];COL[X]=COL[X&LT;&LT;1]+COL[X&LT;&LT;1|1];} struct Nd{int u;ll v;nd (int u,ll v): U (U), V (v) {};nd () {};}; nd query (int tl,int tr,int l,int r,int x) {if (tl<=l&&tr>=r) return nD (Col[x],sm[x]), LL Tm=0;int tc=0,mid= (l+r) >>1;nd o;if (tl<=mid) o=query (Tl,tr,lson), tc+=o.u,tm+=o.v;if (tr >mid) O=query (Tl,tr,rson), tc+=o.u,tm+=o.v;return nd (TC,TM);} void print (int l,int r,int x) {printf ("%d%d%d\n", l,r,sm[x]), if (l==r) return, int mid= (l+r) >>1;print (Lson);p rint ( Rson);} int main () {int n=read (), M=read (), U,v,d,cnt=n;rep (i,1,n) A[i]=b[i]=read (), Rep (i,1,m) {ns[i].u=read (); if (ns[i].u!=3) Ns[i].v=read (), A[++CNT]=B[CNT]=NS[I].V;} Sort (b+1,b+cnt+1); Cnt=unique (b+1,b+cnt+1)-b-1;rep (i,1,n) A[i]=lower_bound (B+1,b+cnt+1,a[i])-b;rep (i,1,n) Update ( a[i],1,b[a[i]],1,cnt,1), Ct[a[i]]++;ll tm;nd o;tm=0;rep (j,2,cnt) {o=query (1,j-1,1,cnt,1); tm+= (LL) ct[j]* (LL) b[j]* O.U-O.V);} Rep (i,1,m) {u=ns[i].u;if (u==3) printf ("%lld\n", TM), else if (u==1) {v=lower_bound (B+1,B+CNT+1,NS[I].V)-b;ct[v]++; Update (v,1,ns[i].v,1,cnt,1), if (v==1) o=nd (0,0), Else O=query (1,v-1,1,cnt,1); tm+= (LL) b[v]*o.u-o.v+ (sm[1]-o.v-(LL) b [V]*ct[v])-(LL) (Col[1]-o.u-ct[v]) *b[v];} Else{v=lower_bound (B+1,B+CNT+1,NS[I].V)-b;if (!ct[V]) puts ("1"); Else{ct[v]--, update (v,-1,-ns[i].v,1,cnt,1), if (v==1) o=nd (0,0); else O=query (1,v-1,1,cnt,1); tm-= (LL) B[v]*o.u-o.v+ (sm[1]-o.v-(LL) b[v]*ct[v])-(LL) (Col[1]-o.u-ct[v]) *b[v];}} printf ("->_->\n");p rint (1,cnt,1);} return 0;}

　　

1394 difference and problem base time limit: 1 seconds space limit: 131072 KB score: 80 Difficulty: 5-level algorithm topic collection attention

There is a multiset s (that is, the elements inside can have duplicates), and in the initial state there are n elements that do the following:

1. Add an element with a value of V to S. Input format is 1 V

2. Remove an element with a value of V inside S. Input format is 2 v

3. Ask for the sum of the absolute value of element 22 within S. Input format is 3

For examples,

Operation 3,|1-2|+|1-3|+|2-3|=4

After Operation 1 4, the number in the collection is 1 2 3 4

Operation 3,|1-2|+|1-3|+|2-3|+|1-4|+|2-4|+|3-4|=10

After Operation 2 2, the number in the collection is 1 3 4

Operation 3,|1-3|+|1-4|+|3-4|=6

Input

The first line enters two integers n,q indicates the number of initial elements and the number of operations in the collection. (1<=n,q<=100,000) The second line gives n integers a[0],a[1],a[2],..., a[n-1], which represents the elements in the initial collection. (0<=a[i]<=1,000,000,000) Next Q line, one operation per line. (0<=v<=1,000,000,000)

Output

For class 2nd operations, if an element with a value of V does not exist in the collection for deletion, output-1. For class 3rd operations, output the answer.

Input example

3 51 2 331 432 23

Output example

4106

51nod1394 Differences and problems